1. NOTES: 14.2 – Gas Laws

2. Pressure-Volume Relationship: (Boyle’s Law) ● Pressure and volume are inversely proportional ● As volume increases, pressure decreases ● As volume decreases, pressure increases ● P 1 V 1 = P 2 V 2

4. Boyle’s Law Example #1: A sample of gas occupies 12.0 L under a pressure of 1.2 atm. What would its volume be if the pressure were increased to 3.6 atm? (assume temp is constant) ● P 1 V 1 = P 2 V 2 ● (1.2 atm)(12.0 L) = (3.6 atm)V 2 ● V 2 = 4.0 L

5. Boyles’s Law Example #2: ● For every 10 meters a scuba diver descends into the water the pressure increases by 1 atmosphere (1 atm or 101.3 kPa). If a scuba diver fills a balloon with 5.00 L of air at the surface and descends to a depth of 100. meters, what will the new volume of the balloon be? (assume pressure at the surface is 1.0 atm!!)

6. Boyles’s Law Example #2: ● For every 10 meters a scuba diver descends into the water the pressure increases by 1 atmosphere (1 atm or 101.3 kPa). If a scuba diver fills a balloon with 5.00 L of air at the surface and descends to a depth of 100. meters, what will the new volume of the balloon be? ● P 1 V 1 = P 2 V 2 ● (1.0 atm)(5.0 L) = (11.0 atm)V 2 ● V 2 = 0.455 L

7. Boyle’s Law Example #3: A high-altitude balloon contains 30.0 L of helium gas at 103.0 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 0.247 atm? (assume constant temperature) **careful with pressure units! ● P 1 V 1 = P 2 V 2 ● (103.0 kPa)(30.0 L) = (25.0 kPa)V 2 ● V 2 = 124 L

8. Temperature-Volume Relationship: (Charles’s Law) ● Volume and temperature are directly proportional ● As temperature increases, volume increases ● As temperature decreases, volume decreases ● (V 1 /T 1 ) = (V 2 /T 2 ) ● ALWAYS USE KELVIN TEMPERATURES!!!!!

10. Charles’ Law Example #1: A sample of nitrogen gas occupies 117 mL at 100°C. At what temperature would it occupy 234 mL if the pressure does not change? (express answer in K and °C) ● V 1 / T 1 = V 2 / T 2 ● (117 mL) / (373 K) = (234 mL) / T 2 ● T 2 = 746 K ● T 2 = 473 ºC

11. Charles’s Law Example #2: ● An 8.00 L capacity Ziplok bag is partially filled with 2.50 L of CO 2 at 0.50 o C. If the bag is set in a sunny window and heated to 70.0 o C, what will be the new volume of the CO 2 ? Will the Ziplok bag burst?

12. Charles’s Law Example #2: ● An 8.00 L capacity Ziplok bag is partially filled with 2.50 L of CO 2 at 0.50 o C. If the bag is set in a sunny window and heated to 70.0 o C, what will be the new volume of the CO 2 ? Will the Ziplok bag burst? ● V 1 / T 1 = V 2 / T 2 ● (2.50 L) / (273.5 K) = (V 2 ) / (343 K) ● V 2 = 3.14 L ● No, the bag won’t burst.

13. Charles’ Law Example #3: A 5.00 L sample of air at -50.0˚C is warmed to 100.0˚C. What is the new volume if the pressure remains constant? ● V 1 / T 1 = V 2 / T 2 ● (5.00 L) / (223 K) = (V 2 ) / (373 K) ● V 2 = 8.36 L

14. Temperature-Pressure Relationship: (Gay-Lussac’s Law) ● Pressure and temperature are directly proportional ● As temperature increases, pressure increases ● As temperature decreases, pressure decreases ● (P 1 /T 1 ) = (P 2 /T 2 ) ● USE KELVIN TEMPERATURES!!!!!

15. Gay-Lussac’s Example #1: The gas in a used aerosol can is at a pressure of 103 kPa at 25.0˚C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928˚C? ● P 1 / T 1 = P 2 / T 2 ● (103 kPa) / (298 K) = (P 2 ) / (1201 K) ● P 2 = 415 kPa

16. Gay-Lussac’s Example #2: The pressure in an automobile tire is 1.85 atm at 27.0˚C. At the end of a trip on a hot sunny day, the pressure has risen to 235 kPa. What is the temperature (in ˚C) of the air in the tire? ● P 1 / T 1 = P 2 / T 2 ● (1.85 atm) / (300. K) = (2.32 atm) / (T 2 ) ● T 2 = 376 K ● T 2 = 103˚C